Transcript solutions for midterm exam

solution for midterm exam
Ch 01 – Ch 06
1. Calculate each value requested for the
following set of scores (2 points each)
X: 1, 3, 0, 2 and Y: 5, 1, –2, –4
a.
b.
c.
d.
e.
ΣX = 6
ΣY = 0
ΣXY = 0
ΣX2 = 1+9+0+4= 14
Σ(Y – 2)2 = 9+1+16+36= 62
2.For the distribution shown in the following table:
(3 points each)
X
25-29
20-24
15-19
10-14
5-9
a.
b.
c.
d.
f
4
6
7
5
3
cf
25
21
15
8
3
c%
100
84
60
32
12
Find the percentile rank for X = 14.5. (32%)
Find the 60th percentile. (19.5)
Find the percentile rank for X = 11. (18%)
Find the 66th percentile. (20.75)
2. c-d
c. (11-9.5)/(14.5-9.5) = (x-0.12)/(0.32-0.12)
1.5/5=(x-0.12)/0.2 0.3*0.2=x-0.12
0.06+0.12=x x = 0.18
d. (y-19.5)/(24.5-19.5)=(0.66-0.6)/(0.84-0.6)
(y-19.5)/5=0.06/0.24y-19.5=5*0.25=1.25
 y = 19.5+1.25 = 20.75
3. (5 points each)
a. A set of n = 6 scores has a mean of M = 10. Another set of
scores has n = 4 and M = 15. If these two sets of scores
are combined, what is the mean for the combined group?
(n=10, ΣX = 120, M = 12)
b. A sample of n = 7 scores has a mean of M = 6. If one score
with a value of X = 12 is removed from the sample, what is
the mean for the remaining scores? (ΣX = 42 ΣX’ = 4212=30M’=30/6=5)
c. A sample of n = 6 scores has a mean of M = 10. One new
score is added to the sample and the new mean is
computed to be M = 9. What is the value of the score that
was added to the sample? (ΣX = 60 ΣX’ =60+x=9*7, X=3)
d. For a sample with M = 40 and s = 4, the middle 95% of the
individuals will have scores between what scores?
M2s=(32,48)
4. (5 points each)
a. A population has µ = 50 and σ = 5. If 10 points
are added to every score in the population, then
what are the new values for the mean and
standard deviation? (µ = 50+10 and σ = 5)
b. A population of scores has µ = 50 and σ = 5. If
every score in the population is multiplied by 3,
then what are the new values for the mean and
standard deviation? (µ = 50*3 and σ = 5*3)
c. Using the definitional formula, compute SS,
variance and the standard deviation for the
following sample of scores. Scores: 3, 6, 1, 6, 5,
3 (M=4, SS = 20; s 2 = 4; s = 2)
5. (9 points each)
a. A population of scores with µ = 73 and σ = 20 is
standardized to create a new population with µ
= 50 and σ = 10. What is the new value for each
of the following scores from the original
population? Scores: 63, 65, 77, 83
b. For a distribution of scores, X = 40 corresponds
to a z-score of z = +1.00, and X = 28 corresponds
to a z-score of z = –0.50. What are the values for
the mean and standard deviation for the
distribution? (Hint: Sketch a distribution and
locate each of the z-score positions.)
5.
a. X=63z=(63-73)/20=−0.5(X’-50)/10= − 0.5
X’=45
X=65z=(65-73)/20= − 0.4(X’-50)/10= − 0.4X’=46
X=77z=(77-73)/20= 0.2(X’-50)/10= 0.2X’=52
X=83z=(83-73)/20= 0.5(X’-50)/10= 0.5X’=55
b.
(40 − µ)/ σ = 1  µ + σ = 40
(28 − µ)/ σ = −0.5  µ − 0.5 σ = 28
1.5 σ = 40-28 = 12  σ =8  µ = 40-8 = 32
6. A normal distribution has a mean of µ = 100 with σ =
20. Find the following probabilities: (3 points each)
use z table
a. p(X > 102)=p(z>102-100/20) =p(z>0.1)=0.5-0.0398=0.4602
b. p(X < 65)=p(z<65-100/20)=p(z<-1.75)=0.5-0.4599=0.0401
c. p(X < 130)= p(z<130-100/20)=p(z<1.5)=0.5+0.4332=0.9332
d. p(95 < X < 105)= p(95-100/20<z<105-100/20)=p(-0.25<z<0.25)
=2*0.0987 = 0.1974
e. What z-score separates the highest 30% from the rest of the
scores? p(0<z<z0)=0.1985≈0.2z0=0.52
7. In an ESP experiment subjects must predict whether a
number randomly generated by a computer will be odd or even.
(hint: probability of odd or even is the same as the probability of
head or tail by tossing a fair coin) (5 points each)
a. What is the probability that a subject would
guess exactly 18 correct in a series of 36 trials?
With n = 36 and p = q = 1/2, you may use the normal approximation
with µ = np = 18 and  2 = npq = 9   = 3.
X = 18 has real limits of 17.5 and 18.5 corresponding to z = 0.17
and z = +0.17.
p(x=18) = p (-0.17<z<0.17)=2*p(0<z<0.17)=2*0.0675= 0.135
7b.
b. What is the probability that a subject would
guess more than 20 correct in a series of 36
trials?
X > 20  X > 20.5  z > 20.5-18/3=0.83
p(X>20.5) = p(z>0.83) = 0.5-p(0<z<0.83) = 0.5 - 0.2967 =
0.2033